What is the value of this expression when c = -4 and d = 10?

A footbridge is 5 feet wide and is built with wood planks that are 6 inches wide. How many planks wide is the path?

Ede4ka [16]

Here we have to know the conversion between feet and inches.

1 Feet = 12 inches (This is given and is to be known)

So the 5 feet wide footbridge when converted to inches would be about 5*12 = 60 inches

The width of the footbridge in inches = 60 inches

Each wooden plank is 6 inches,

So the number of wooden planks that are required to build the foot bridge of 60 inches = 60/6 = 10 planks

Hence, the path path is 10 planks wide

7 0

3 years ago

Find the unit rate. 180 miles in 3 hours The unit rate is miles per hour. Pls hurry

Natali [406]

Answer:

60 miles per hour

Step-by-step explanation:

Consider the structure of the wording: "miles per hour"

The word "per" means for each. There is you clue.

Each is, one of and one of is unit measurement

So the controlling element in this question is that you need to convert the 3 hours into 1 hour.

Using ratio properties

180

3

=

x

miles

1

hour

We need to 'force' the left hand side into the same form as the right. That is: we need to convert the denominator into 1 and see what happens to the numerator.

⇒

180

÷

3

÷

3

=

x

1

⇒

60

1

=

x

1

So the unit rate of speed is

60

miles

per

−−−

hour

4 0

3 years ago

What's the answer for this question? (The numbers after the letters are indexes btw) 27a9 x 18b5 x 4c2 Over 18a4 x 12b2 x 2c

zysi [14]

Answer:

$\frac{27a^9 * 18b^5 * 4c^2 }{18a^4 * 12b^2 * 2c} = \frac{9}{2}a^5b^3c$

Step-by-step explanation:

Given

$\frac{27a^9 * 18b^5 * 4c^2 }{18a^4 * 12b^2 * 2c}$

Required

Simplify

$\frac{27a^9 * 18b^5 * 4c^2 }{18a^4 * 12b^2 * 2c}$

Cancel out 18

$\frac{27a^9 * b^5 * 4c^2 }{a^4 * 12b^2 * 2c}$

Divide 4 and 2

$\frac{27a^9 * b^5 * 2c^2 }{a^4 * 12b^2 *c}$

Divide 27 and 12 by 3

$\frac{9a^9 * b^5 * 2c^2 }{a^4 * 4b^2 *c}$

Apply law of indices

$\frac{9a^{9-4} * b^{5-2} * 2c^{2-1} }{4}$

$\frac{9a^5 * b^3 * 2c }{4}$

Divide 2 and 4

$\frac{9a^5 * b^3 * c}{2}$

$\frac{9a^5b^3c}{2}$

Rewrite as:

$\frac{9}{2}a^5b^3c$

Hence:

$\frac{27a^9 * 18b^5 * 4c^2 }{18a^4 * 12b^2 * 2c} = \frac{9}{2}a^5b^3c$

4 0

3 years ago

Factor 26r3s + 52r5 – 39r2s4.

algol [13]

Answer:

13r²(2rs + 4r³ - 3s⁴)

Step-by-step explanation:

In equation 26r³s + 52r⁵ - 39r²s⁴;

The GCF of 26, 52, and 39 = 13

The GCF of r³, r⁵ and r² = r²

The GCF of s, (no "s"), and s⁴ = no "s" (Since one of the number doesn't have "s")

Now we can factor out 13r² from all three expressions;

26r³s + 52r⁵ - 39r²s⁴

=> 13r²(2rs) + 13r²(4r³) - 13r²(3s⁴)

To factor it all together;

13r²(2rs + 4r³ - 3s⁴)

Hope this helps!

7 0

3 years ago

Please answer this asap.

vodomira [7]

The answer would be 39.78

4 0

3 years ago

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